Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $k = \dfrac{5y^2 - 30y}{5y^3 - 15y^2 - 50y} \times \dfrac{2y^2 - 14y - 36}{y - 9} $
Solution: First factor out any common factors. $k = \dfrac{5y(y - 6)}{5y(y^2 - 3y - 10)} \times \dfrac{2(y^2 - 7y - 18)}{y - 9} $ Then factor the quadratic expressions. $k = \dfrac {5y(y - 6)} {5y(y + 2)(y - 5)} \times \dfrac {2(y + 2)(y - 9)} {y - 9} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {5y(y - 6) \times 2(y + 2)(y - 9) } { 5y(y + 2)(y - 5) \times (y - 9)} $ $k = \dfrac {10y(y + 2)(y - 9)(y - 6)} {5y(y + 2)(y - 5)(y - 9)} $ Notice that $(y + 2)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {10y\cancel{(y + 2)}(y - 9)(y - 6)} {5y\cancel{(y + 2)}(y - 5)(y - 9)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $k = \dfrac {10y\cancel{(y + 2)}\cancel{(y - 9)}(y - 6)} {5y\cancel{(y + 2)}(y - 5)\cancel{(y - 9)}} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $k = \dfrac {10y(y - 6)} {5y(y - 5)} $ $ k = \dfrac{2(y - 6)}{y - 5}; y \neq -2; y \neq 9 $